If x and y are positive, is x < y?

Hi KarishmaB MartyMurray chetan2u
I have attached my solution on paper. For statement 2, is it correct to stop where I figure out it could either be x+y <6 & x>y or x+y >6 & x<y. So, I have two possibilities of relationship between x and y. Hence insufficient. Thanks for your feedback on my solution.

Yes, in this question, you could stop at the point you have two different solutions as there are no restrictions on x and y except they are positive integers.

Use the number line approach for stmnt 2 to avoid writing anything.


(1) \(\sqrt{x} < \sqrt{y}\)
Squaring both sides because principal square roots are always positive, we get
x < y
Sufficient.


(2) \((x-3)^2 < (y-3)^2\)
Since both sides are squares so positive, we can take the square root to get

|x - 3| < |y - 3|
(Check this post if you are not sure how: https://anaprep.com/algebra-squares-and-square-roots/ )

Distance of x from 3 < Distance of y from 3

So imagine the number line. Either is possible.

______ 1 (= y)_________ 2 (=x) _________ 3 ________ 4 (=x) _______ 5 (= y) _______


x has a smaller range around 3 than y, but in some cases x < y (e.g. x = 2 and y = 5) while in other cases y < x (e.g. x = 2, y = 1)

Not Sufficient

Answer (A)

Check: How to use number line on GMAT: https://youtu.be/3gxVx3Y9xJA